From an Eigenvalue to an Eigenspace
An eigenvalue describes a scalar action of a matrix on certain nonzero vectors. There may be many such vectors for the same eigenvalue: every nonzero vector in the corresponding eigenspace is an eigenvector. Geometric multiplicity measures the size of that space by its dimension. In practice, this turns an eigenvalue question into a homogeneous-system calculation.
Let \(\mathbb{F}\) be \(\mathbb{R}\) or \(\mathbb{C}\), let \(\mathbf{A}\in\mathbb{F}^{n\times n}\), and let \(\lambda\in\mathbb{F}\). An eigenvector for \(\lambda\) is a nonzero vector \(\mathbf{x}\) satisfying \(\mathbf{A}\mathbf{x}=\lambda\mathbf{x}\). The eigenspace \(E_{\lambda}\) consists of the zero vector together with all eigenvectors for \(\lambda\). Rearranging the eigenvalue equation gives \((\mathbf{A}-\lambda\mathbf{I}_n)\mathbf{x}=\mathbf{0}\), so the eigenspace is the null space of \(\mathbf{A}-\lambda\mathbf{I}_n\).
The word “dimension” is important: it counts the vectors in a basis, not the total number of eigenvectors. An eigenspace of positive dimension contains infinitely many vectors over either \(\mathbb{R}\) or \(\mathbb{C}\), but its dimension is a finite integer between \(1\) and \(n\). The lower bound holds because \(\lambda\) is an eigenvalue, so its eigenspace contains a nonzero vector.
Reading the Dimension from Row Reduction
To calculate geometric multiplicity, form \(\mathbf{A}-\lambda\mathbf{I}_n\), row-reduce it, and solve the resulting homogeneous system. Pivot variables are determined by the free variables. Each free variable can be assigned independently, and the resulting solution vectors form a basis for the null space.
Proof. Write \(\mathbf{R}=\operatorname{rref}(\mathbf{A}-\lambda\mathbf{I}_n)\). The row operations used to obtain \(\mathbf{R}\) preserve the solutions of the homogeneous system: swapping equations does not change them, multiplying an equation by a nonzero scalar is reversible, and adding a multiple of one equation to another is reversible. Thus \(\operatorname{Nul}\mathbf{R}=E_{\lambda}\). Let the free variables be \(t_1,\dots,t_q\). In each nonzero row of \(\mathbf{R}\), its pivot variable is determined by a linear expression in the free variables. Consequently every solution has the form \(t_1\mathbf{z}_1+\cdots+t_q\mathbf{z}_q\), where \(\mathbf{z}_j\) is the solution obtained by setting \(t_j=1\) and the other free variables to \(0\). These vectors span the null space. They are linearly independent because the free-variable coordinates of \(\sum_{j=1}^q c_j\mathbf{z}_j\) are \(c_1,\dots,c_q\); if that vector is zero, all \(c_j=0\). Hence the \(\mathbf{z}_j\) form a basis of the eigenspace, and its dimension is \(q\). There are \(n\) variable columns altogether, so \(q=n-r\). \(\square\)
The proof gives more than a count: it explains how to produce an eigenspace basis. When checking the result, substitute each proposed basis vector into \(\mathbf{A}\mathbf{x}=\lambda\mathbf{x}\). That verifies that the vectors really are eigenvectors; the free-variable argument verifies that they span all of them.
Worked Example: An Eigenspace of Dimension Two
Find the geometric multiplicity and a basis for the eigenspace of \(\lambda=3\) for
Subtract \(3\) from each diagonal entry to form \(\mathbf{A}-3\mathbf{I}_3\), then eliminate the second row using \(R_2\to R_2-2R_1\):
The homogeneous equation is \(x_1+2x_2+3x_3=0\). The variables \(x_2\) and \(x_3\) are free. Set \(x_2=s\) and \(x_3=t\); then \(x_1=-2s-3t\), so
Thus a basis is \(\left(\begin{bmatrix}-2&1&0\end{bmatrix}^{\top},\begin{bmatrix}-3&0&1\end{bmatrix}^{\top}\right)\), and \(\operatorname{gm}(3)=2\). The eigenvalue condition can also be checked directly. For \(\mathbf{v}_1=(-2,1,0)^{\top}\), \(\mathbf{A}\mathbf{v}_1=(-6,3,0)^{\top}=3\mathbf{v}_1\). For \(\mathbf{v}_2=(-3,0,1)^{\top}\), \(\mathbf{A}\mathbf{v}_2=(-9,0,3)^{\top}=3\mathbf{v}_2\). The vectors are independent because their second and third coordinates force both coefficients in any zero combination to be zero.
Worked Example: An Eigenspace of Dimension One
Find the geometric multiplicity and eigenspace for \(\lambda=2\) when
Subtract \(2\mathbf{I}_3\). Swap the second and third rows, then scale the second row by \(\tfrac13\):
The equations are \(x_2=0\) and \(x_3=0\), leaving \(x_1\) free. Therefore \(E_2=\operatorname{span}\{(1,0,0)^{\top}\}\) and \(\operatorname{gm}(2)=1\). Check: \(\mathbf{A}(1,0,0)^{\top}=(2,0,0)^{\top}=2(1,0,0)^{\top}\). Every solution of the reduced system is a multiple of this vector, so it spans the whole eigenspace.
Worked Example: A Parameter Changes the Eigenspace Dimension
For real \(t\), determine the geometric multiplicity of \(\lambda=1\) for
For every \(t\), the first standard basis vector is an eigenvector, since \(\mathbf{A}_t(1,0,0)^{\top}=(1,0,0)^{\top}\). Thus \(1\) is indeed an eigenvalue for all parameter values. Subtracting the identity gives
If \(t\ne0\), scale the second row by \(1/t\). The reduced equations give \(x_2=0\) and \(x_3=0\), so the eigenspace has basis \(\{(1,0,0)^{\top}\}\) and geometric multiplicity \(1\). If \(t=0\), the second row is zero, leaving \(x_1\) and \(x_3\) free. The eigenspace then has basis \(\{(1,0,0)^{\top},(0,0,1)^{\top}\}\) and geometric multiplicity \(2\).
For \(t=0\), the second basis vector checks as \(\mathbf{A}_0(0,0,1)^{\top}=(0,0,1)^{\top}\); the first checks as \(\mathbf{A}_0(1,0,0)^{\top}=(1,0,0)^{\top}\). For \(t\ne0\), the first vector still checks as above, and the reduced equations show that every eigenvector lies in its span.
Why the Dimension Does Not Depend on Coordinates
A matrix can change when the coordinates used to describe a linear map change. The eigenspace itself is still a vector space, and its dimension does not depend on which basis is used to represent it. Similar matrices encode the same linear map in different coordinates; the following result makes the invariance precise.
Proof. For every \(\mathbf{x}\in E_{\lambda}(\mathbf{A})\), compute \((\mathbf{B}-\lambda\mathbf{I}_n)\mathbf{P}^{-1}\mathbf{x} =\mathbf{P}^{-1}(\mathbf{A}-\lambda\mathbf{I}_n)\mathbf{x} =\mathbf{0}\). Thus \(\mathbf{P}^{-1}\mathbf{x}\in E_{\lambda}(\mathbf{B})\). Since \(\mathbf{P}^{-1}\) is invertible, it sends nonzero vectors to nonzero vectors, so this also shows that \(\lambda\) is an eigenvalue of \(\mathbf{B}\). Conversely, the same calculation with \(\mathbf{A}=\mathbf{P}\mathbf{B}\mathbf{P}^{-1}\) shows that \(\mathbf{P}\) sends \(E_{\lambda}(\mathbf{B})\) into \(E_{\lambda}(\mathbf{A})\). These maps are inverses on the two eigenspaces. In particular, applying \(\mathbf{P}^{-1}\) to a basis of \(E_{\lambda}(\mathbf{A})\) gives a basis of \(E_{\lambda}(\mathbf{B})\): it spans because the map is onto, and it is independent because the map is one-to-one. The dimensions are therefore equal. \(\square\)
Common Pitfall: Counting Pivots Instead of Free Variables
In the first worked example, the RREF has one pivot and two free-variable columns. A common error is to report geometric multiplicity \(1\) because there is one pivot. That counts the number of pivot columns, which determines how many variables are constrained, not the null-space dimension. The correct calculation is \(3-1=2\), or, directly, the two free variables give the two basis vectors found in the example.
Another mistake is to count eigenvectors rather than basis vectors. In that same example, the zero vector and every vector \(s\mathbf{v}_1+t\mathbf{v}_2\) lie in the eigenspace. Counting these vectors does not give its dimension. The dimension is \(2\) because \(\mathbf{v}_1,\mathbf{v}_2\) form a basis: they are independent and span the entire eigenspace. The zero vector belongs to every eigenspace, but by definition it is not an eigenvector.
Geometric multiplicity is a null-space calculation for a specified eigenvalue. The algebraic multiplicity, by contrast, is defined using how many times \(\lambda\) occurs as a root of the characteristic polynomial. These are different definitions and involve different computations. To find geometric multiplicity, solve the homogeneous system and count its free variables; do not substitute a characteristic-polynomial root count for that process.
Check Your Understanding
For each question, identify the null-space calculation involved and justify the dimension from the solution set.
- For \(\mathbf{A}=\begin{bmatrix}3&0&0\\0&3&1\\0&0&3\end{bmatrix}\), find a basis for \(E_3\) and calculate \(\operatorname{gm}(3)\).
- If \(\operatorname{rref}(\mathbf{A}-\lambda\mathbf{I}_5)\) has three pivot columns and \(\lambda\) is an eigenvalue, what is \(\operatorname{gm}(\lambda)\)? Explain using free variables.
- For \(\mathbf{A}_t=\begin{bmatrix}2&0&0\\0&2&t\\0&0&2\end{bmatrix}\), determine the geometric multiplicity of \(2\) for \(t=0\) and \(t\ne0\).
- Explain why the zero vector is in \(E_{\lambda}\) but is not counted as an eigenvector.
- Suppose \(\mathbf{B}=\mathbf{P}^{-1}\mathbf{A}\mathbf{P}\) with \(\mathbf{P}\) invertible. Describe the map that carries \(E_{\lambda}(\mathbf{A})\) to \(E_{\lambda}(\mathbf{B})\).
